In C.D.Sogge's Fourier Integrals in Classical Analysis pp.128-129, he proved Lemma4.2.3(Tauberian Lemma):
Lemma. Let$g(\lambda)$ be a piece-wise continuous tempered function of $\mathbb{R}$. Assume that for $\lambda>0$ $$|g(\lambda+s)-g(\lambda)|\leq C(1+\lambda)^a,0<a\leq1$$ Then if $\hat{g}(t)=0$ when $|t|\leq1$(The hat means Fourier transform) we have: $$|g(\lambda)|\leq C(1+\lambda)^a$$#
by using an identity:
$$|G(\lambda)|=|(G'*\psi)(\lambda)|\leq C(1+\lambda)^a\int|\psi(s)|(1+|s|)^ads\leq C(1+\lambda)^a$$
My confusion is that when the $\hat{\psi}(t):=\frac{\eta(t)}{it}$ where $\eta\in\mathcal{S}$ satisfying $\eta(t)=0$ when $|t|\leq\frac{1}{2}$ and $\eta(t)=1$ when $|t|>1$. It could be that $\psi$ is not integrable at all since $\frac{1}{t}$ is not necessarily integrable at infinity. So how could this identity holds?
What is more, in the same book, pp.127 in the proof to Theorem4.2.1 the same argument arise. Altough this can be argued using oscillatory integral in Chap1.
Remarks from Professor It could be found in Hormander's PDO Vol.3 this Lemma holds under a more restrictive assumption, and it suffices for later applications.
My question is whether the identity above can be argued correctly? If so, what technique should be used?(I will be deeply appreciated if a detail explanation is given.)
Thanks.